What is the integral of ((x^2-1))/(x+1) ?

The integral of ((x^2-1))/(x+1) is (x^2)/2-x+C

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integral of ((x^2-1))/(x+1)

\int \frac{( x ^{2} - 1 )}{x + 1} d x

\frac{x ^{2}}{2} - x + C

Solution steps

\int \frac{x ^{2} - 1}{x + 1} d x

Simplify

\frac{x ^{2} - 1}{x + 1} : x - 1

= \int x - 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x d x - \int 1 d x

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

= \frac{x ^{2}}{2} - x

Add a constant to the solution

= \frac{x ^{2}}{2} - x + C

What is the integral of ((x^2-1))/(x+1) ?
The integral of ((x^2-1))/(x+1) is (x^2)/2-x+C